a)\(\sqrt{3x}=\sqrt{5}\)

\(\Leftrightarrow3x=5\Leftrightarrow x=\dfrac{5}{3}\)

b)\(\sqrt{x-5}=\sqrt{3x+2}\)

\(\Leftrightarrow x-5=3x+2\)

\(\Leftrightarrow-2x=7\Leftrightarrow x=-\dfrac{7}{2}\)

a)\(=3\left(x-5\right)^2\)

b)\(=x\left(y-2x-1\right)\)

c)\(=\left(x+1\right)\left(x-8\right)\)

d)\(=4x\left(x+y+2\right)\left(x+y-2\right)\)

e)\(=\left(y+z\right)\left(x-2\right)\)

g)\(=\left(x+3+y\right)\left(x+3-y\right)\)

Từ \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(x+y\right)\end{matrix}\right.\)

Và \(C=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{z}\)

\(=\dfrac{-z}{y}\cdot\dfrac{-x}{z}\cdot\dfrac{-y}{z}=-\left(\dfrac{xyz}{xyz}\right)=-1\)

các biểu thức trong căn pt hết về HĐT rồi phá ra là done

bài 2: \(x=y=-1|;x=-1;y=0|;x=1;y=-2|;x=y=1\)- khao học đã chứng minh

bình phuognw 2 vé rồi thu gọn là được

a)\(x^3 - 3x^2 + 1 - 3x\)

\(=(x+1)(x^2-4x+1)\)

b)\(2x^2 - 18\)

\(=2(x-3)(x+3)\)

c)\(x^2 - 7xy + 10y^2\)

\(=(2y-x)(5y-x)\)

d)\(x^4 - 7x^2 + 1\)

\(=(x^2-3x+1)(x^2+3x+1)\)

e)\(x^2 + 8x + 7\)

\(=(x+1)(x+7)\)